What is the molar concentration of K4Fe(CN)6 solution if 40 ml were used to titrate 150 mg of Zn ( that are dissolved ) forming K2Zn3[Fe(CN)6]2 ?

Question

I really need help for this excersice , I don't know what to do...
Note: In my book the answer is 0,0382 M

Answer 1

First, your post is confusing. I found the following equation on the net and I assume the 150 mg Zn is in the form of the ion and not solid. If all of that is true, then

3Zn(NO3)2 + 2K4[Fe(CN)6] → K2Zn3[Fe(CN)6]2 + 6KNO3

Then mols Zn = grams/atomic mass = ?
Convert mols Zn to mols K4[Fe(CN)6]
Looking at the coefficients in the balanced equation, mols K4Fe(CN)6 = mols Zn x (2/3)
Then M K4Fe(CN)6 = mols/L = ?
You might want to confirm that equation. I believe that will give you 0.0382 M Post your work if you get stuck.